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Lecture 4 One Dimensional Kinematics.
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Lecture 4 One Dimensional Kinematics General Physics I, Lec4 By/T.A.Eleyan
In lecture this we discuss motion in one dimension. We introduce definitions for displacement, velocity and acceleration, and derive equations of motion for bodies moving in one dimension with constant acceleration. We apply these equations to the situation of a body moving under the influence of gravity alone. One Dimensional Kinematics** **One dimensional kinematics refers to motion along a straight line. General Physics I, Lec4 By/T.A.Eleyan
Kinematicsis that branch of physics which involves the description of motion, without examining the forces which produce the motion. Dynamics, on the other hand, involves an examination of both a description of motion and the forces which produce it. General Physics I, Lec4 By/T.A.Eleyan
Distance and displacement • Distance is the total length of travel. • It is always positive. • It is measured by the meter. Displacement is defined as the change in position of an object. • xf= final value of x, xi = initial value of x • Change can be positive, negative or zero. • Displacement is a vector ‘D’ (Delta)=change General Physics I, Lec4 By/T.A.Eleyan
Distance or Displacement? Distance may be, but is not necessarily, the magnitude of the displacement Displacement Distance General Physics I, Lec4 By/T.A.Eleyan
Example, starting with xi = 60 m and ending at xf = 150 m, the displacement is Δ x = xf - xi = 150 m - 60 m = 90 m xi < xf . Δx > 0 since Example, starting with xi = 150 m and ending at xf = 60 m, the displacement is Δx = xf - xi = 60 m - 150 m = -90 m Δx < 0 sincexi > xf . General Physics I, Lec4 By/T.A.Eleyan
Average Speed and Velocity Speed and velocity are not the same in physics! Speed is rate of change of distance: (always positive) Velocity is rate of change of displacement: (positive, negative or zero) velocity is a vector Here we are just giving the ‘x-component’ of velocity, assuming the other components are either zero or irrelevant to our present discussion General Physics I, Lec4 By/T.A.Eleyan
The Position - Time graph The average velocity between two times is the slope of the straight line connecting those two points. average velocity from 2 to 3 sec is negative average velocity from 0 to 3 sec is positive General Physics I, Lec4 By/T.A.Eleyan
Instantaneous Velocity The velocity at one instant in time is known as the instantaneous velocity and is found by taking the average velocity for smaller and smaller time intervals: The speedometer indicates instantaneous velocity (Dt 1 s). On an x vs t plot, the slope of the line tangent to the curve at a point in time is the instantaneous velocity at that time. General Physics I, Lec4 By/T.A.Eleyan
The average velocity of a particle is defined as the ratio of the displacement to the time interval. The instantaneous velocity of a particle is defined as the limit of the average velocity as the time interval approaches zero. General Physics I, Lec4 By/T.A.Eleyan
Example: A Particle movies along the x-axis. According to the expression 1- determine the displacement of the particle in the time interval t=0s to t=1s x(0)=0 and x(1)=-2 m 2- Calculate the average velocity in the time interval t=0 to t=1 3- Find the instantaneous velocity of the particle at t=2.5 second At t=2.5s , General Physics I, Lec4 By/T.A.Eleyan
Acceleration Often, velocity is not constant, rather it changes with time. The rate of change of velocity is known as acceleration. This is the average acceleration. • Acceleration is a vector • The unit of acceleration is: m/s2 positive, negative or zero General Physics I, Lec4 By/T.A.Eleyan
Instantaneous Acceleration If we wish to know the instantaneous acceleration, we once again let t 0: General Physics I, Lec4 By/T.A.Eleyan
The Velocity - Time graph Graphically, acceleration can be found from the slope of a velocity vs. time curve. For these curves, the average acceleration and the instantaneous acceleration are the same, because the acceleration is constant. General Physics I, Lec4 By/T.A.Eleyan
Example A car moves from a position of +4.0 m to a position of –1.0 m in 2.0 sec. The initial velocity of the car is –4.0 m/s and the final velocity is –1 m/s. (a) What is the displacement of the car? (b) What is the average velocity of the car? (c) What is the average acceleration of the car? Answer: (a) Dx = xf – xi = –1.0 m – (+4.0 m) = – 5 m (b) vav = Dx/Dt = (– 5.0 m)/(2.0 s) = – 2.5 m/s (c) General Physics I, Lec4 By/T.A.Eleyan
Example: The velocity of a particle moving according to the expression 1) Find the average acceleration in the time interval t=0 to t=2s 2) Determine the acceleration at t=2s At, t=2 then General Physics I, Lec4 By/T.A.Eleyan